The big idea: You can't weigh a gas easily, but you can measure its volume. Two tools turn a gas volume into an amount in moles.
- The ideal gas equation, PV = nRT, works at any pressure and temperature. - The molar volume, Vm = 22.7 dm³ mol⁻¹, is a shortcut that works only at STP (standard temperature and pressure).
What STP means: STP = standard temperature and pressure = 273 K (0 °C) and 100 kPa.
At STP, one mole of any ideal gas takes up the same volume: 22.7 dm³. So at STP you don't need PV = nRT — just divide the volume by 22.7.
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At STP, the identity of the gas doesn't matter — equal volumes hold equal numbers of moles. So the amount of a gas is simply its volume divided by the molar volume, Vm = 22.7 dm³ mol⁻¹.
- amount of gas (mol)
- volume of gas at STP (dm³)
- molar volume at STP = 22.7 dm³ mol⁻¹
Watch the volume unit: Vm is in dm³ mol⁻¹, so the volume must be in dm³.
- 1 dm³ = 1000 cm³, so divide cm³ by 1000 to get dm³. - 1 dm³ = 1 litre, so 22.7 dm³ = 22.7 L.
Worked example — volume of gas at STP
Calculate the volume, at STP, of 0.150 mol of carbon dioxide, CO2.
Solution
- Formula first — rearrange n = V/Vm for the volume:
- Substitute (Vm = 22.7 dm³ mol⁻¹ at STP):
- Work it out — keep the unit:
Final answer
V = 3.41 dm³ (at STP).
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When conditions are not STP, use the full ideal gas equation. It links pressure, volume, amount and temperature. Rearrange it for whatever you are asked to find.
- pressure (Pa)
- volume (m³)
- amount of gas (mol)
- gas constant = 8.31 J K⁻¹ mol⁻¹
- temperature (K)
SI units — the make-or-break step: PV = nRT only works with SI units, because R = 8.31 J K⁻¹ mol⁻¹ is in SI.
- P in pascals (Pa) — multiply kPa by 1000. - V in cubic metres (m³) — divide dm³ by 1000, or cm³ by 10⁶. - T in kelvin (K) — add 273 to °C.
Get a unit wrong and the answer is off by a factor of 1000 or more.
Worked example — moles from P, V and T
A flask of volume 2.00 dm³ holds nitrogen gas at a pressure of 120 kPa and a temperature of 25 °C. Calculate the amount, in moles, of nitrogen. (R = 8.31 J K⁻¹ mol⁻¹.)
Solution
- Convert to SI units first. Pressure: 120 kPa → Pa:
- Volume: 2.00 dm³ → m³ (÷1000):
- Temperature: 25 °C → K (+273):
- Formula first — rearrange PV = nRT for n:
- Substitute and solve:
- Work it out:
Final answer
n = 0.0969 mol (≈ 0.097 mol).
How this is tested: S1.5 gas calculations turn up as a Paper 1A one-step MCQ and as a marked Paper 2 calculation.
The two recurring asks are: 'find the volume of gas at STP' (use Vm = 22.7) and 'find the moles of a vaporized compound from P, V and T' (use PV = nRT). Often the answer then feeds into a molar-mass step.
The marks live in unit conversion — Pa, m³ and K for PV = nRT; dm³ for Vm.
Three easy marks: (1) Decide first: STP given → Vm; otherwise → PV = nRT. (2) Convert every value to the right unit before substituting. (3) Write the rearranged formula before the numbers, and carry the unit.
IB-style question — volume of hydrogen at STP
Zinc reacts with excess hydrochloric acid to produce 0.0800 mol of hydrogen gas, H2. Calculate the volume of hydrogen produced at STP. [1]
Solution
- STP is stated, so use the molar volume — formula first:
- Substitute (Vm = 22.7 dm³ mol⁻¹):
- Work it out:
Final answer
V = 1.82 dm³ at STP.
IB-style question — molar mass via the gas equation
When 0.512 g of a volatile liquid is fully vaporized, it occupies 195 cm³ at a pressure of 99.0 kPa and a temperature of 90.0 °C. Calculate the amount, in moles, of vapour, then its molar mass. (R = 8.31 J K⁻¹ mol⁻¹.) [2]
Solution
- Convert to SI units. P = 99.0 kPa, V = 195 cm³, T = 90.0 °C:
- Formula first — rearrange PV = nRT for n:
- Amount of vapour:
- Now use M = m/n to get the molar mass:
- Work it out:
Final answer
n = 6.40 × 10⁻³ mol; M = 80.0 g mol⁻¹.